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I'm a physics student getting a bit out of my depth in maths, and I'd really like someone to confirm that my reasoning below is correct.

We're working in the Hilbert space $\mathcal{H}=L^2(\mathbb{R},dx)$. Define the operator with its maximal domain,

$$ A := \frac{d^2}{dx^2} + \frac{1}{4}x^2 , $$ $$ \mathcal{D}(A) = \{ \varphi \in \mathcal{H} | A\varphi \in \mathcal{H} \}. $$

$A$ is symmetric by a trivial by-parts calculation. To confirm that it's self-adjoint, we check the deficiency index. This means looking at

$$ n_\pm = \text{dim Ker}(A^\dagger \mp i). $$

This means solving the differential equation

$$ \frac{d^2 f}{dx^2} + \frac{1}{4} x^2 f(x) \mp if(x) =0 $$

which gives (according to Mathematica)

$$ f(x) = c_1 D_{-3/2}((-1)^{1/4} x) + c_2 D_{1/2}((-1)^{3/4} x) $$

in the "$-if$" case and

$$ f(x) = c_1 D_{1/2}((-1)^{1/4} x) + c_2 D_{-3/2}((-1)^{3/4} x) $$

in the "$+if$" case, where $D_n(z)$ is the parabolic cylinder function. These four solutions certainly aren't square-integrable, and I am fairly sure that $c_1$ and $c_2$ cannot be set to make either of the solutions square-integrable either, as I think that the parabolic cylinder functions diverge by different positive powers at infinity. Therefore there is no solution in $\mathcal{H}$, hence $n_-=n_+=0$, hence $A$ is Self-Adjoint. This means that I can apply the spectral theorem, which means that I can write any function $\psi(x) \in \mathcal{H}$ as

$$ \psi(x) = \sum_{k=1,2}\int_\mathbb{R} da \; \psi_k(a) f_k(a;x) $$

where $f_k(a;x)$, $k=1,2$ are two linearly independent solutions of

$$ \frac{d^2 f}{dx^2} + \frac{1}{4} x^2 f(x) - af(x) =0. $$

Is this all correct? It very much seems correct to me but I've learned half the definitions I've used for it this week so I'm not at all confident, and I really need to be able to rely on the result!

J_B_Phys
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  • If your Hilbert space is real, what's the difference between symmetric and self-adjoint? Also, can you explain what deficiency index means? The handful of search results I've skimmed don't actually define it. – anon Dec 02 '20 at 01:24
  • Oh, by symmetric I mean "Hermitian". The difference between Hermitian and Self-Adjoint only comes about with unbound operators on infinite-dimensional Hilbert Spaces, as unbound operators can't be defined on the whole Hilbert space. Self-Adjoint is the requirement that an operator adjoint's domain be the same as the operator's domain, as well as the requirement that the operators be equal within their mutual domain. And I think I've technically defined the deficiency index well in the question? The crucial result is just that if A is Hermitian then it's also self adjoint iff $n_+=n_-=0$. – J_B_Phys Dec 02 '20 at 01:31
  • (Hermitian being only the requirement that they're the same on $\mathcal{D}(A) \subseteq \mathcal{D}(A^\dagger)$. Self-Adjoint is Hermitian requirement plus $\mathcal{D}(A)=\mathcal{D}(A^\dagger)$.) – J_B_Phys Dec 02 '20 at 01:35
  • I don't know anything about the parabolic cylinder function, but essential self adjointness on the span of the Hermite functions can be proved using Carleman's test. (Carleman's test is also proved by showing that a vector orthogonal to the range of $A+i$ cannot be in $L_2$.) – Keith McClary Dec 10 '20 at 17:14

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